Question: Quadrilateral $ABCD$ has right angles at $B$ and $D$, and $AC=3$. If $ABCD$ has two sides with distinct integer lengths, then what is the area of $ABCD$? Express your answer in simplest radical form.
Solution: Triangles $\triangle ABC$ and $\triangle ADC$ are both right and share hypotenuse $AC$, which has length $3$. Thus we have $$AB^2+BC^2 = AD^2+DC^2 = 3^2 = 9.$$The only possible integer values for $AB,$ $BC,$ $AD,$ or $DC$ are $1$ and $2$. Thus we may assume that one leg of $\triangle ABC$ has length $1$ and one leg of $\triangle ADC$ has length $2$ (it doesn't matter if the labels $B$ and $D$ have to be swapped to make this true).

If one leg of $\triangle ABC$ has length $1,$ then the other leg has length $\sqrt{3^2-1^2} = \sqrt{8} = 2\sqrt{2}$. If one leg of $\triangle ADC$ has length $2,$ then the other leg has length $\sqrt{3^2-2^2}= \sqrt{5}$. Thus, quadrilateral $ABCD$ is divided by its diagonal $AC$ into right triangles of area $\frac{1\cdot2\sqrt 2}{2}=\sqrt 2$ and $\frac{2\cdot\sqrt 5}{2}=\sqrt 5$. So, the area of quadrilateral $ABCD$ is $\boxed{\sqrt 2+\sqrt 5}$.